Question

Find $\int si{n}^{4}xdx$

Answer 1

$\int {\sin }^{4}xdx$

$\int {\sin }^{3}x.\sin xdx$

Since ${\sin }^{3}x=3\sin x-4{\sin }^{3}x⟹{\sin }^{3}x=\frac{3\sin x-\sin 3x}{4}$

$\frac{1}{4}\int (3\sin x-\sin 3x)\sin xdx$

$=\frac{1}{4}\int (3{\sin }^{2}x-\sin 3x\sin x)dx$

$=\frac{1}{4}\int \left(3\right(\frac{1-\cos 2x}{2})-\frac{\cos 2x-\cos 4x}{2})dx$

$=\frac{1}{8}(3x-2\sin 2x+\frac{1}{4}\sin 4x)$

$=\frac{3x}{8}-\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x$